Question

An aircraft with mass 43581 kg has a constant acceleration of 1.7 m/s2 as it starts (from rest) down the runway for take-off. The plane reaches take-off speed in 34.8 seconds. What is the aircraft’s take-off speed in miles per hour?

Answer 1

132.337 miles per hour

Step-by-step explanation:

We are given that an aircraft with amss 43581 kg

The acceleration of an aircraft =
`1.7 m/s^2`

Initial velocity of an aircraft =0 (Because it start moving from rest)

Time taken by aircraft to reaches take-off speed=34.8 s

Using newton's second law of motion

`s=ut+(1)/(2)at^2`

Substitute all values then we get

`s=0* 34.8+(1)/(2)1.7(34.8)^2`

`s=1029.384 m`

Using Newton's third law of motion

`v^2-u^2=2as`

Substitute all given values then we get

`v^2-0=2* 1.7* 1029.384`

`v=√(3499.9056)=59.16 m/s`

1 meter=0.000621371 miles

1 hour =3600 seconds

Therefore, v=
`(59.16* 0.000621371)/((1)/(3600))`=132.337 miles per hour